题目1096：日期差值（a-b=(a-c)-(b-c)）

http://ac.jobdu.com/problem.php?pid=1096

题目描述：

有两个日期，求两个日期之间的天数，如果两个日期是连续的我们规定他们之间的天数为两天

输入：

有多组数据，每组数据有两行，分别表示两个日期，形式为YYYYMMDD

输出：

每组数据输出一行，即日期差值

样例输入：

20110412

20110422

样例输出：

11

思路：

计算2011 00 00到2011 04 12的天数为acount；计算2011 00 00到2011 04 22的天数为bcount；最终结果应该是acount-bcount+1。

代码：

# include<iostream>
using namespace std;

int main()
{
    int m[] = { , , , , , , , , , , , ,  };

    char a[], b[];
    int ya, yb, ma, mb, da, db;
    int i, j, k, acount, bcount;
    while (cin >> a >> b)
    {
        ya = (a[] + ( - '')) *  + (a[] + ( - '')) *  + (a[] + ( - '')) *  + (a[] + ( - ''));
        yb = (b[] + ( - '')) *  + (b[] + ( - '')) *  + (b[] + ( - '')) *  + (b[] + ( - ''));
        ma = (a[] + ( - '')) *  + (a[] + ( - ''));
        mb = (b[] + ( - '')) *  + (b[] + ( - ''));
        da = (a[] + ( - '')) *  + (a[] + ( - ''));
        db = (b[] + ( - '')) *  + (b[] + ( - ''));
        //cout << ya << " " << ma << " " << da << endl;
        //cout << yb << " " << mb << " " << db << endl;

        acount = ;
        for (i = ; i < ma; i++)
        {
            acount += m[i];
        }
        if (ma> && ((ya %  ==  && ya %  != ) || ya %  == ))
        {
            acount += ;
        }
        acount += da;

        bcount = ;
        for (i = ya; i < yb; i++)//not <=
        {
            if ((i %  ==  && i %  != ) || i %  == )
            {
                bcount += ;
            }
            else
            {
                bcount += ;
            }
        }
        for (i = ; i < mb; i++)
        {
            bcount += m[i];
        }
        if (mb> && ((yb %  ==  && yb %  != ) || yb %  == ))
        {
            bcount += ;
        }
        bcount += db;

        cout << bcount - acount +  << endl;
    }

    return ;
}
/**************************************************************
    Problem: 1096
    User: mmcNuaa@163.com
    Language: C++
    Result: Accepted
    Time:0 ms
    Memory:1520 kb
****************************************************************/